TITLE: Show $1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$.
QUESTION [10 upvotes]: Show that $$1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$$
where $|q|<1$ (q can be complex number).
The hint is to convert the left side to a double series.
One can see that the nominators on the left side is double, triple, ... of those on the right side, while denominators on the right side are square of those on the left side. Both sides have 1, which seems redundant.

I'm not sure how to proceed from these observations. Any advice will be helpful. I can try to solve it with some suggestion and if I still can't solve it I will then explain what puzzles me. If anyone gives the answer I guess I will look at only part of it and work from that.


Edits:
I post my answer inspired by the comments and some other thoughts about the question below.
Is there any other ways to solve it?

REPLY [3 votes]: These series are not equal for all $|q|<1$. Here's an explanation.
First, let
$$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n},\qquad |bq|<1$$
and $$B(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2},\qquad |bq|<1.$$
Let
$$F_1(q)=A(-2,-1;-q)=\frac{2q}{1-q}+\frac{4q}{1+q^2}+\frac{8q^3}{1-q^3}+...$$
and $$F_2(q)=B(-1,-1;-q)=\frac{q}{(1-q)^2}+\frac{q^2}{(1+q^2)^2}+\frac{q^3}{(1-q^3)^2}+...$$

You assert that $$1+4F_1(q)=1+8F_2(q).\tag{*}$$ This is incorrect. Below is an explanation.


First of all, define the functions $\phi_n,\sigma_n:\Bbb R\times\Bbb R\to \Bbb R$ to satisfy
$$\begin{align}
A(a,b;q)&=\sum_{n\ge1}\phi_n(a,b)q^n\\
B(a,b;q)&=\sum_{n\ge1}\sigma_n(a,b)q^n.
\end{align}$$
Then we can see that
$$\begin{align}
A(a,b;q)&=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}\\
&=\frac1b\sum_{n\ge1}a^n\frac{bq^n}{1-bq^n}\\
&=\frac1b\sum_{n\ge1}a^n\sum_{k\ge1}b^kq^{nk}\\
&=\frac1b\sum_{n\ge1}\sum_{k\ge1}a^nb^kq^{nk}\\
&=\frac1b\sum_{m\ge1}q^m\sum_{nk=m}a^nb^k\\
&=\sum_{m\ge1}q^m\sum_{d|n}a^db^{m/d-1}.
\end{align}$$
Thus $$\phi_n(a,b)=\sum_{d|n}a^db^{n/d-1}.$$
Then, preform the index shift $d=n/j$ for $j|n$:
$$\phi_n(a,b)=\sum_{j|n}a^{n/j}b^{j-1}=\frac{a}{b}\sum_{j|n}b^ja^{n/j-1}=\frac{a}{b}\phi_n(b,a).\tag1$$
Then take $\partial_b=\frac{\partial}{\partial b}$ of $A(a/q,b;q)$:
$$\begin{align}
\partial_b A(a/q,b;q)&=\sum_{n\ge1}\partial_b\frac{a^n}{1-bq^n}\\
&=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2}\\
&=B(a,b;q).
\end{align}$$
Then from $(1)$, we have that $A(a,b;q)=\frac{a}{b}A(b,a;q)$, so that
$$\begin{align}
A(a/q,b;q)&=\frac{a}{bq}A(b,a/q;q)\\
&=a\sum_{n\ge1}\frac{(bq)^{n-1}}{1-aq^{n-1}}\\
&=\frac{a}{1-a}+a\sum_{n\ge1}\frac{(bq)^{n}}{1-aq^{n}}\\
&=\frac{a}{1-a}+bA(a,b;q).
\end{align}$$
Thus,
$$\begin{align}
B(a,b;q)&=\partial_b A(a/q,b;q)\\
&=\partial_b \left\{\frac{a}{1-a}+bA(a,b;q)\right\}\\
&=\partial_b bA(a,b;q)\\
&=A(a,b;q)+b\partial_b A(a,b;q)\\
&=\sum_{n\ge1}q^n\phi_n(a,b)+\sum_{n\ge1}q^nb\partial_b \phi_n(a,b)\\
&=\sum_{n\ge1}q^n\left\{\sum_{d|n}a^db^{n/d-1}+b\sum_{d|n}\left(\tfrac{n}{d}-1\right)a^db^{n/d-2}\right\}\\
&=\sum_{n\ge1}q^n\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.
\end{align}$$
Hence $$\sigma_n(a,b)=\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.$$

Since $F_1(q)=A(-2,-1;-q)$ and $F_2(q)=B(-1,-1;-q)$, we have
$$F_1(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\phi_n(-2,-1)\right)=\sum_{n\ge1}p_1(n)q^n,$$
and
$$F_2(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\sigma_n(-1,-1)\right)=\sum_{n\ge1}p_2(n)q^n.$$
Then your claim $(*)$ is equivalent to $p_1(n)=2p_2(n)$, or
$$\sum_{d|n}(-1)^{d+n/d-1}2^d=2\sum_{d|n}(-1)^{d+n/d-1}\frac{n}{d}.\tag{*'}$$
So $(*)$ is true if and only if $(*')$ is true.
To check this, let
$$f(n)=\sum_{d|n}(-1)^{d+n/d-1}(2^d-2n/d).$$
$(*)$ Is true if and only if $f(n)$ is identical to $0$.

It is quite easy to show that $f(3)=-(2-6)-(2^3-2)=-2$, so $f(n)$ is not identical to $0$ and your claim is false.